The minimum number of NAND gates required to construct an OR gate is

Options

A. 4
B. 6
C. 5
D. 3

Explanation

To determine the minimum number of NAND gates required to construct an OR gate, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the NAND Gate**:
- A NAND gate performs the operation \( Y = \overline{A \cdot B} \). This means it outputs a false signal only when both inputs are true.

2. **Recall De Morgan's Theorems**:
- De Morgan's Theorem states that:
- \( \overline{A \cdot B} = \overline{A} + \overline{B} \)
- \( \overline{A + B} = \overline{A} \cdot \overline{B} \)
- We will use this theorem to express the OR operation using NAND gates.

3. **Express OR in terms of NAND**:
- The OR operation can be expressed as:
- \( A + B = \overline{\overline{A} \cdot \overline{B}} \)
- This means we need to first find \( \overline{A} \) and \( \overline{B} \) (the NOT operation), and then perform the AND operation on these results, followed by a NOT operation.

4. **Constructing the OR gate using NAND gates**:
- To get \( \overline{A} \), we can use a NAND gate with both inputs as A:
- \( \overline{A} = A \text{ NAND } A \)
- To get \( \overline{B} \), we can use a NAND gate with both inputs as B:
- \( \overline{B} = B \text{ NAND } B \)
- Now, we have both \( \overline{A} \) and \( \overline{B} \).

5. **Perform the AND operation on \( \overline{A} \) and \( \overline{B} \)**:
- We can use another NAND gate to perform the AND operation:
- \( \overline{A} \cdot \overline{B} = \overline{(\overline{A} \text{ NAND } \overline{B})} \)
- This requires one more NAND gate.

6. **Final NOT operation**:
- Finally, we need to apply a NAND operation to get the final output:
- \( Y = \overline{(\overline{A} \cdot \overline{B})} \)
- This requires one more NAND gate.

### Summary of Gates Used:
- 1 NAND gate for \( \overline{A} \)
- 1 NAND gate for \( \overline{B} \)
- 1 NAND gate for \( \overline{A} \cdot \overline{B} \)
- 1 NAND gate for the final output

### Total:
Thus, the minimum number of NAND gates required to construct an OR gate is **4**.